# Finding a k value which will satisfy 1 positive and 1 negative solution for a given function

Let $f(x) = x^2-10x+9$. The function has $x$ intercepts at $1$ and $9$. For which values of $k$ does $f(x) = 2x + k$ have $1$ positive root and $1$ negative root?

My try: I got $f(x) = x^2-12x + 9 = k$. But now I noticed that I cannot factorise the expression. Can somebody please help me answer this question

• From your attempt at factoring, are you looking for only integer roots? Nov 11, 2014 at 17:06

Well, first of all, $k>9$. First, place $k$ to the left side:

$$x^2-12x+9-k$$

To find the roots, you will need to use the quadratic formula:

$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

With the following variables:

$a=1$, $b=-10$ and $c=9-k$. You will get the following using the quadratic formula:

$$\frac{12\pm\sqrt{(-12)^2+4\cdot1\cdot(9-k)}}{2\cdot1}$$

Simplifying the equation:

$$x =\frac{12\pm2\sqrt{27+k}}{2} = 6\pm\sqrt{27+k}$$

To get the roots, ${27+k}$ needs to be larger than $36$ (because $6-\sqrt{36} = 0$).

So, $27 +k>36 \iff k>9$

• Why do we need integer roots? Why wouldn't any $k > 9$ satisfy the problem? Nov 11, 2014 at 16:07
• Sorry, I thought we were looking for integer root. Thanks for the notification! Nov 11, 2014 at 16:49

Consider $x^2-12x+9-k$. When $|x|$ is large, the quadratic is positive. So all you need is to ensure that when $x=0$, it is negative. Invariably it will have to cross the axes on either side. Now can you find what values $k$ could take?

Can you also argue why if this quadratic has a non-negative value when $x=0$, it cannot have two roots of opposite signs?